More Trig

Tickle

New member
Joined
Apr 27, 2015
Messages
9
Thanks Soroban.

I have another one here.


[ ( 2 + tan^2x ) / ( sec^2x ) ] - 1



I think it starts out with:


[ ( 2 + ( sinx / cosx )^2 ) / ( 1 / cosx )^2 ] - 1

then

[ 2 + ( sinx / cosx )^2 * ( cosx / 1 )^2 ] - 1

Then I don't know where to go with it. I get lost.

Kind Regards,
Tickle
 
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Thanks Soroban.

I have another one here.


[ ( 2 + tan^2x ) / ( sec^2x ) ] - 1



I think it starts out with:


[ ( 2 + ( sinx / cosx )^2 ) / ( 1 / cosx )^2 ] - 1

then

[ {2 + ( sinx / cosx )^2} * ( cosx / 1 )^2 ] - 1 ...... missing parentheses

Then I don't know where to go with it. I get lost.

Kind Regards,
Tickle

Please tell us:

\(\displaystyle \displaystyle{\frac{1}{sec(x)} \ = \ ??}\)

\(\displaystyle \displaystyle{{tan(x)}*{cos(x)} \ = \ ??}\)
 
Last edited by a moderator:
I'm not sure how you guys are making nice equations on the board. So I will attach the equation I did in microsoft word.

Well I guess I can't even upload a .docx. jeez.

< image link removed >

Kind Regards,
Tickle
 
Last edited by a moderator:
I'm not sure how you guys are making nice equations on the board.
By using LaTeX (pronounced "LAY-teck"). You can find a quickie reference here.

So I will attach the equation I did in microsoft word.

\(\displaystyle \dfrac{2\, +\, \tan^2(x)}{\sec^2(x)}\, -\, 1\, =\)

\(\displaystyle \dfrac{2\, +\, \left( \dfrac{\sin(x)} {\cos(x)} \right)^2} {\left( \dfrac{1} {\cos(x)} \right)^2} \, -\, 1\, =\)

\(\displaystyle 2\, +\, \left(\dfrac{\sin(x)}{\cos(x)}\right)^2\, \cdot\, \left(\dfrac{\cos(x)}{1}\right)\, -\, 1\, =\)

Then I'm not sure where to go with this.
I've formatted what you'd posted. Note that, as posted, you've changed the expression between the second and third lines of computations, because you left out the grouping, putting the "2+" as a separate thing (rather than being "joined" with the tangent fraction). Also, the square in the one denominator has disappeared. As a result, you've posted this:

\(\displaystyle 2\, +\, \left[\left(\dfrac{\sin(x)}{\cos(x)}\right)^2\, \cdot\, \left(\dfrac{\cos(x)}{1}\right)\right]\, -\, 1\, =\)

But you can't flip an expression that merely contains fractions. You have to have a fractional term. So the first step would be to back up to the second line:

. . . . .\(\displaystyle \dfrac{2\, +\, \left( \dfrac{\sin(x)} {\cos(x)} \right)^2} {\left( \dfrac{1} {\cos(x)} \right)^2} \, -\, 1\)

...and then multiply out squares and convert to fractional-only form before flipping:

. . . . .\(\displaystyle \dfrac{\left(\dfrac{2\, \cos^2(x)}{\cos^2(x)}\, +\, \dfrac{\sin^2(x)}{\cos^2(x)}\right)}{\left(\dfrac{1}{\cos^2(x)}\right)}\, -\, 1\)

. . . . .\(\displaystyle \dfrac{\left(\dfrac{2\, \cos^2(x)\, +\, \sin^2(x)}{\cos^2(x)}\right)}{\left(\dfrac{1}{\cos^2(x)}\right)}\, -\, 1\)

. . . . .\(\displaystyle \left(\dfrac{2\, \cos^2(x)\, +\, \sin^2(x)}{\cos^2(x)}\right)\left(\dfrac{\cos^2(x)}{1}\right)\, -\, 1\)

...and so forth. ;)
 
Here is what I have so far. Not sure what to do next.

< image link removed >
 
Last edited by a moderator:
Hello Tickle:

I have removed your image links from this thread because they lead to a web page that contains objectionable images (i.e., scantily-clad women).

If you desire to use third-party image-hosting web sites here -- instead of uploading directly to our server -- you must choose a host that does not append objectionable images to their pages.

Thank you ~ Mark :cool:
 
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